Covered in lectures. Check back once the chapter is concluded.
2 Complex-valued functions
Definition 2.1 A complex function \(f\colon D\to\C\) is a map with domain of definition \(D\subset\C\) and codomain the complex plane. Thus, \(f\) assigns to each \(z=x+iy\in D\) in the domain a complex number\[f(z)=u(z)+iv(z).\]We call \(u\colon D\to\R\) the real part and \(v\colon D\to\R\) the imaginary part of the complex function \(f.\)
Example 2.1
Covered in lectures. Check back once the chapter is concluded.
Example 2.2
Covered in lectures. Check back once the chapter is concluded.
We have already met several complex functions in the previous section.
Example 2.3
Covered in lectures. Check back once the chapter is concluded.
Example 2.4
Covered in lectures. Check back once the chapter is concluded.
Definition 2.2 We define the following subsets of the complex plane: \[\begin{align*} \C^\t&=\{z\in\C\mid z\neq0\} &&\text{\textbf{punctured plane}}\\ \C^-&=\C\setminus\{x\in\R\mid x\leqslant 0\} &&\text{\textbf{slit plane}}\\ S&=\{z=x+iy\mid y\in(-\pi,\pi)\} &&\text{\textbf{principal strip}}\\ \H&=\{z=x+iy\mid y>0\}&&\text{\textbf{upper half-plane}} \end{align*}\]
Covered in lectures. Check back once the chapter is concluded.
Example 2.5
Covered in lectures. Check back once the chapter is concluded.
Proposition 2.1
The exponential function is surjective onto \(\C^\t.\)
We have \[\exp(\log(w))=w\ (\forall w\in\C^-) \tag{2.1}\] \[\log(\exp(z))=z\ \; (\forall z\in S). \tag{2.2}\]
Hence the restriction \(\exp|_S\) of the exponential to the principal strip is a bijection \(\exp|_S\colon S\to\C^-\) onto the slit plane, with inverse \(\log(w).\)
Proof.
Covered in lectures. Check back once the chapter is concluded.
Since the graph \(\Ga(f)=\{(z,w)\in D\t\C\mid f(z)=w\}\) of a complex function is a subset of four-dimensional space, we cannot visualize complex functions as easily as real functions. We will now discuss some alternatives.
Image grid
A useful way to picture a complex function is to sketch its values on a grid \(G.\) The image grid \(f(G)\) is a distorted version of the original grid which can be navigated easily using the grid lines. For example, to determine \(f(1+2i),\) take one step in \(x\)-direction and two steps in \(y\)-direction on the distorted grid. Formally, let \(G=\{z=x+iy\in\C\mid x\in\Z\text{ or }y\in\Z\}\) be the unit grid and define the image grid as (see Figure 2.1)
\[f(G)=\{w=u+iv\in\C\mid\exists z\in G: f(z)=w\}.\]
In practice, the image grid can often be described by finding a familiar equation that all of its members \(u+iv\) satisfy. When this is not possible, a computer will help sketching an approximate image grid.
Example 2.6
Covered in lectures. Check back once the chapter is concluded.
Example 2.7
Covered in lectures. Check back once the chapter is concluded.
The previous example can be generalized.
Example 2.8
Covered in lectures. Check back once the chapter is concluded.
Domain colouring
We represent each unit complex number \(e^{i\th}\) by a color on the color wheel. The modulus \(r\) of an arbitrary complex number \(re^{i\th}\) will be represented by the lightness of the color. This assigns a unique color to each complex number, see Figure 2.2. Pure white is never used and would correspond to infinity. Pure black corresponds to the origin.
This is less useful for calculations by hand but generates artistic images using a computer.
This can be used for visualizing complex functions. Draw each point \(z\) in the domain of \(w=f(z)\) using the color for \(w.\)
3-dimensional graphs
Another approach is to plot the 3-dimensional graph of any of the following real-valued functions \(D\to \R\) \[u, v, |f|=\sqrt{u^2+v^2}.\]
Again, the missing information can be color-coded (see Figure 2.4).
Questions for further discussion
- What are the functions in Example 2.1 geometrically?
- Does Equation 2.2 remain valid for all \(z\in\C\)? What is the correct modification?
- The two zeros in Figure 2.3 have a slightly different character. What is the difference between the zeros that might account for this?
- In (ex-Ch2_Fig_Image_Grid_1?) and (ex-Ch2_Ex1z?)} almost all the image gridlines meet at a right angle. The only exception is at \(f(0)\) in Example 2.6. Use polar coordinates to explain this behavior of \(f(z)=z^2\) at \(z=0.\)
- Is it always possible to extend the domain of definition of a complex function? Is this always sensible?
- Can you think of other branches of the logarithm?
2.1 Exercises
Describe the image set of the complex function \(f(z)=\frac{1+z}{1-z}\) with domain \(D=\C\setminus\{1\}.\) In other words, determine the set of all \(w\in\C\) for which \(w=\frac{1+z}{1-z}\) has a solution \(z\in D.\)
Sketch the following curves \(z(t)\) in the complex plane, where \(t\) is a real parameter.
- \(z(t)= t(1+i)\) for \(0\leqslant t\leqslant 1\)
- \(z(t)= \cos(t)+i\sin(t)\) for \(0\leqslant t\leqslant \pi\)
- \(z(t)= \cos(t)-i\sin(t)\) for \(0\leqslant t \leqslant \pi/2\)
- \(z(t)=\frac{1}{1+it}\) for \(t\in\R\)
Let \(\log(z)\) be the principal branch of the logarithm. Compute \[\log(2i), \log(1+i), \log(-3i), \log(5).\]
Describe the following complex functions geometrically. \[f(z)=3z, f(z)=iz, f(z)=\frac{(1+i)}{\sqrt{2}}z\]
Determine a domain of definition for the following complex functions. \[f(z)=\frac{1}{z}, f(z)=\frac{1+z}{z-1}, f(z)=\frac{z^2-4}{z^2+2z}, f(z)=\frac{1}{\exp(z)}. \]
Determine the domain of definition for \(f(z)=\frac{1}{\sin(z)}.\)
Find all solutions to the following equations:
- \(e^z=-1\),
- \(\sin(z)=-i\)
Prove that the composition \(f\circ f'\) of two Möbius transformations is again a Möbius transformation. If we associate to \(f, f'\) the matrices \[A=\begin{pmatrix} a&b\\ c&d \end{pmatrix}, A'=\begin{pmatrix} \alpha &\beta\\ \gamma& \delta \end{pmatrix} \in \M_{2\times 2}(\C),\]
show that \(f\circ f'\) is associated to the matrix product \(AA'.\)
Draw the image grid for \(\exp\colon\C\to\C^\t.\)
Further resources
- https://youtu.be/NtoIXhUgqSk 5 ways to visualize a complex function
- Animation of Möbius transformations: https://youtu.be/0z1fIsUNhO4 and https://www-users.cse.umn.edu/~arnold/moebius/